The bore with corners and the passage beside it

Two islands in one film bored a round shaft with two round holes and found what a second hole adds that a bigger hole does not: in Prandtl’s soap-film picture of torsion, every hole is an island floating at its own height, set by a balance between the film’s pull round its edge and the pressure on it, and two islands talk to each other through the film between them. The steel between two holes is a thin strip of film between two plateaus, and it is steep.

That essay ended on bores that are not round. Hollow shafts are bored with splines, hexagonal sockets, square drives and keyed holes, because the bore is often how the torque gets in. A polygonal bore’s island floats at a height set by the same balance, round an edge that has corners, and the film between it and the outside surface falls across steel whose thickness varies round the bore. This essay asks what the corners of the bore do, and where a passage drilled through the wall — for oil, for a pin, for a sensor’s cable — should be placed relative to them.

The film round a polygonal bore #

Prandtl’s analogy states torsion as a soap film stretched over the section’s outline and pushed up by a uniform pressure: the film’s height is the stress function, its slope is the shear stress, and twice the volume under it is the torque per unit twist. The first essay on the analogy set that out; the island the film floats added the rule for a hole. A hole is a region where there is no steel to carry stress, so the film over it must be flat — an island — and it floats at whatever height makes the film’s pull round the island’s edge balance the pressure on its area:

$$\oint_{\text{edge}} \frac{\partial \phi}{\partial n}\, ds = -2A_{\text{hole}}.$$

A round bore centred on the axis floats at the height a solid shaft’s film already has at that radius, so the film is unchanged outside it and the bore carries the solid shaft’s stress at its own radius: for an 8 mm bore in a 25 mm shaft, 0.33 of the surface stress. A hexagonal bore with its flats 8 mm from the axis has the same island height near enough, but its edge is no longer a contour of the solid film. The film has to bend round six corners.

At each corner of the bore, the steel has a re-entrant corner: the material wraps 240° round the hole’s 120° corner. The corner a soap film cannot finish found the same geometry at the root of a keyway, on the shaft’s outside, and found that a film cannot finish a re-entrant corner: with a sharp corner its slope is unbounded. Here the corners face inwards, and the bore’s fillet is what makes the stress at them a number.

The stress round the bore rises from 0.28 at the middle of a flat to 0.53 at a corner rounded to 1.5 mm, nearly twice. The flats carry less than a round bore would, because a flat is nearer the axis than a round bore through the corners would be and the film there is shallow. The corners carry much more. Round a polygonal bore the stress gathers at the corners, and with corners rounded to 1.5 mm the corner of a socket in the middle of a shaft carries more than half the stress the shaft carries at its outside surface.

How much a fillet buys #

The fillet is the designer’s only lever on the corner, and it works the way it does at a keyway’s root: a corner’s stress falls roughly as the fillet’s radius to a small negative power, so doubling the fillet buys back a share of the excess rather than halving it. From 1.5 mm to 3 mm the corner falls from 0.53 to 0.45; at 6 mm it is 0.38, near the round bore’s 0.33, and the bore has become a hexagon with very round corners. A socket that has to take a hexagonal key cannot have corners rounded that far, because the key’s corners have to fit; the fillet a drive socket can have is set by the key it drives.

The numbers at the smallest fillet carry the method’s limit. The film is solved on a grid about a third of a millimetre square, and a 1.5 mm fillet spans five cells. Solved on a finer grid the 1 mm fillet’s corner moves by about five per cent, which is the accuracy to read the corner values with; the flats, the passage and the island heights are far less sensitive.

Fewer sides, sharper corners #

The number of sides decides two things at once: how sharp each corner is, and how far it reaches out from the axis. A triangle with its flats 8 mm from the axis has its corners 16 mm out, so they reach into film twice as steep, and turn through 60°, which makes the re-entrant corner in the steel 300°. Its corners carry 1.39 of the plain shaft’s surface stress — more than the shaft’s own surface carries — while its flats, close to the axis, carry almost nothing. A square drive’s corners carry 0.80, a hexagon’s 0.53, a twelve-sided spline’s 0.42. As the sides increase the polygon becomes a circle and both lines converge on the round bore’s 0.33.

That is why splines have many teeth and why square drives are given generous corner radii on shafts that carry their full torque: the shape of the bore can move the governing stress from the outside of the shaft to the inside, where it cannot be seen, polished or rolled.

A passage beside the bore #

A passage drilled along the shaft through its wall — for oil to a bearing, a pin, a cable — is a second island. Two islands in one film found that two round holes interact through the film between them, the steel between them steepening as they approach. Beside a polygonal bore the interaction depends on what part of the bore the passage faces.

Opposite a corner the two islands share the thinnest steel there is: the ligament between the passage and the bore’s corner is 1.4 mm. The film has to fall from the passage’s island to the bore’s in that distance, and both edges steepen. The bore’s corner goes from 0.54 to 1.15 — more than doubled — and the passage peaks at 1.25. Opposite a flat, the passage faces the bore across 2.5 mm of steel to a flat whose film is shallow, and the bore does not notice it: its corners stay at 0.54. The passage itself peaks at 1.19, a little less than opposite the corner.

Turning the passage from a corner to a flat lowers the bore’s peak steadily, by half, and the passage’s own peak slightly. There is no bearing at which the passage is better placed than opposite a flat, and none at which it is worse than opposite a corner. Turned from a corner to a flat, the passage stops loading the bore.

How near the bore #

The instinct with a hole near another hole is to move it away. In a shaft that instinct points the wrong way, because the torsional stress grows with the radius: a passage moved outwards from the bore moves into steeper film.

Opposite a flat the passage is best as near the bore as it will go: at 12.3 mm from the axis it peaks at 1.13 and leaves the bore at 0.63. Moved outwards it climbs steadily — 1.4 at 16 mm, 2.0 at 20 mm — and near the surface, where the steel between the passage and the outside is under a millimetre thick, it peaks at 2.84. Opposite a corner the picture near the bore is different: at 12.3 mm the passage and the bore both reach 1.48, because they share a ligament under a millimetre thick; moved out by a millimetre and a half the bore drops to 0.97 while the passage rises, and beyond 14 mm the two orientations give the same passage stress and differ only at the bore.

The best passage in this shaft still peaks at 1.13 times the plain shaft’s surface stress, above the 1.01 the hollow shaft carries at its surface. A passage anywhere in the wall governs the shaft. What its placement decides is by how much, and whether it drags the bore’s corners up with it.

A square drive leaves less room #

The hexagon is a forgiving bore: its corners are only 15 per cent further from the axis than its flats, so a passage opposite a corner still has a millimetre or two of steel to work with. A square drive’s corners reach 34 per cent further out, and the choice of bearing stops being a refinement. With the same 5 mm passage 14 mm from the axis, opposite a corner of a square bore whose flats are 8 mm out, the ligament to the corner is 0.8 mm and both edges reach 2.05 times the plain shaft’s surface stress — the bore’s corner two and a half times its own value without the passage. Turned 45° to face a flat, the passage peaks at 1.23 and the bore’s corners stay at 0.76, slightly below the 0.80 they carry alone, because the passage takes a little of the film’s slope away from the nearest corners as it steepens its own. At the same 14 mm a hexagonal bore gives 1.31 and 0.95 opposite a corner and 1.28 and 0.57 opposite a flat. The fewer the bore’s sides, the more the passage’s bearing matters, and on a square drive it decides whether the bore’s corners are the shaft’s weakest point or nearly unchanged.

What the corners cost in stiffness, and what they cost in stress #

The corners are expensive in stress and almost free in stiffness, which is why they are easy to overlook. Twice the volume under the film is the torque per unit twist, and the corners change only a thin rim of the film near the axis, where it is lowest. The hexagonal bore leaves the shaft with 0.981 of the plain shaft’s torsional constant, against 0.985 for a round bore of the flats’ radius — the hexagon removes a tenth more steel and costs the stiffness less than half a per cent. A square bore leaves 0.977 and a triangle 0.939, the triangle’s corners reaching out into steel that matters. The passage costs more than the bore’s shape does: 0.969 with it 13 mm out opposite a flat, 0.945 with it near the surface. A shaft checked for its twist would pass with any of these bores. A shaft checked for its stress is governed by the corners or the passage.

In a working shaft the numbers become stresses directly. A 50 mm shaft carrying 1 kN·m has a plain surface stress of $2T/\pi R^3 = 2 \times 10^6/(\pi \times 25^3) = 40.7$ N/mm². The hexagonal bore’s 1.5 mm corners then carry 21.6 N/mm², its flats 11.4. A passage opposite a flat, close to the bore, peaks at 46 N/mm²; the same passage near the surface at 116; the same passage opposite a corner, close to the bore, leaves 60 at both its own edge and the bore’s corner. For a shaft whose torque reverses millions of times, a notch’s fatigue strength depends on that elastic peak, and the difference between 46 and 116 N/mm² is the difference between a passage that is harmless and one that sets the shaft’s life.

A hole in a twisted wall, by hand #

The passage’s own numbers follow from one classical result. A small round hole in a field of antiplane shear — the stress state of a shaft’s wall in torsion — doubles the shear stress at its edge. The nominal stress at radius $r$ in a solid shaft is $\tau_0 r/R$, so a small hole centred at $r_p$ peaks at about twice the nominal stress at its outer edge, $2\tau_0 (r_p + a)/R$.

For the passage 13 mm out with a 2.5 mm radius, that is $2 \times 15.5/25 = 1.24$, against the film’s 1.19 opposite a flat — close, the film a little lower because the bore takes some of the stress round to its far side. At 16 mm it is $2 \times 18.5/25 = 1.48$ against the film’s 1.4. Near the surface the rule fails, as it should: at 21.7 mm the hole’s edge is 0.8 mm from the surface, the steel between them is no longer a wide field round a small hole, and the film’s 2.84 is the thin ligament’s own steepness. The rule gives the shape of the curve; the film gives the two ends where the hole is not alone.

Which free body the balance comes from #

The island’s height is the only part of the solution that is not local, and it comes from a free body. Cut the shaft along the boundary of the bore, all the way along its length, and take the tube of steel inside the cut — which is to say, nothing, because the bore is empty. The shear flow round the bore’s edge must therefore circulate without any net twist of the boundary relative to the steel around it: the film’s pull round the edge balances the pressure on the island’s area, which is the statement that the cut has no torque passing through the hole. The island’s height is the one number per hole that makes that true, and it is what makes every hole talk to every other: a passage beside the bore changes the film round the bore’s edge, and the bore’s island moves to restore its balance.

What a film over one section cannot see #

The shaft is in pure torsion. A drive shaft also bends, and in bending a passage near the outside of the section is in the most stress for a different reason. The best place for a passage under combined loading is a compromise the torsion figures alone do not make.

The bore carries no load. A socket that takes its torque from a key is loaded on its flats by the key’s contact pressure, which concentrates at the flats’ edges near the corners; the stress here is the shaft’s response to a torque applied at its ends, not to a torque delivered through its bore.

The corners are resolved to about five per cent. The film is solved on a grid a third of a millimetre square; the flats, the passage and the islands are converged closely, the corner values at the smallest fillets less so.

The passage runs the length of the shaft. The film is a two-dimensional solution, exact for a section far from any end and any change of section. A passage drilled radially to meet an axial one, or one that stops part way along, has three-dimensional ends where the film’s stress flows round the end of the hole, and the concentration there is higher than anywhere the section can show — which is why oil holes are given generous entry chamfers, and why the section’s numbers are the stress along the hole rather than at its mouth.

Everything is elastic. In a ductile shaft the corners and the passage’s edge yield locally and shed their peaks, which matters for a single overload and not for fatigue, where the elastic peak is what a crack starts from — as it did at the keyway’s root.

Still open: the bore that is a spline #

A hexagon has six corners; a spline has twenty or thirty, each a tooth root with its own small fillet, and between them the teeth are slender cantilevers that carry the mating shaft’s load into the bore. In torsion alone the spline’s film is nearly a round bore’s at the tip circle with a ring of shallow notches, and the analogy predicts a small peak at each root. Whether a splined bore’s roots behave like many small corners — each a little worse than a round bore, as the trend from six sides to twelve suggests — or whether the teeth, carrying their own share of the torque in contact with the mating spline, put a load into each root that the film of an empty bore cannot see, is the question that turns this bore with corners into the bore that actually transmits the torque, and it is where the film with two islands meets a contact problem it was never asked to solve.